you smuggled an extra i into your first expression (force of habit, I guess). With e^(pi/2) *i^i instead of e^(i*pi/2) *i^i, the whole thing resolves to
e^(Pi/2-Pi/2 - 2*Pi*n)=e^(-2Pi*n)=1 for all whole numbers n

No one is going to point out that the caption says "Mathematicians are no longer allowed to sporting events"?
Sorry guys. I was a math major but I was also a writing minor. Awkward wording pops out to me. At me. Whatever you prefer.

It is multi-valued.
e^(Pi/2) i^i = e^(Pi/2) e^(i ln i) = e^(Pi/2) e^(i (ln |i| + i arg(i))) = e^(Pi/2) e^(- Pi/2 - 2 n Pi) = e^(- 2 n Pi)
for any integer n. There is no 'i' term in the exponent, so this is not a single value. For n=0, we get 1 (and indeed, this would be the value associated with the principal branch of the complex logarithm), but it's just as reasonable to claim that this sign says "We're #e^(16 Pi)", (approximately 6.7 x 10^21)

Although it's a reasonable gripe that i^i has infinitely many possible values, I'm more concerned that the guy in the middle thinks 0^0 is 1. I think that maybe the mathematicians had a buddy from the physics department come along.

The value of 0^0 is "undefined". It's a classic trick question about double limits (limits of limits). It isn't 1: think about the limit of the limit of f(u,v) = u^v as both u and v go to 0. If you take the limit as u goes to 0 first (with v not 0) then that limit is 0. Do it the other way around and take the limit as v goes to 0 first and you get 1. So you can approach 0^0 two different ways and get two different answers for what it should be. If 0^0 really had a value, you couldn't get two different limits for it.